Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = (int)1e5 + 7;
const int LOG = 18;
const int INF = (int)1e6;
struct node {
  vector<int> grl;
  int sz;
  node() {
    grl.resize(11, INF);
    sz = 0;
  }
  void insert(int a) {
    if (!binary_search(grl.begin(), grl.end(), a)) {
      grl[10] = a;
      sort(grl.begin(), grl.end());
      sz = (sz < 10) ? sz + 1 : 10;
    }
  }
  void print(int k) {
    cout << min(sz, k);
    for (int i = 0; i < min(sz, k); i++) {
      cout << " " << grl[i];
    }
    cout << endl;
  }
};
node upg[MAXN][LOG + 1];
int up[MAXN][LOG + 1];
int tin[MAXN];
int tout[MAXN];
int timer = 0;
int n;
int m;
int q;
vector<int> g[MAXN];
node gc[MAXN];
node merge(node a, node b) {
  for (int i = 0; i < b.sz; i++) {
    a.insert(b.grl[i]);
  }
  return a;
}
void dfs(int v, int p = 1) {
  tin[v] = ++timer;
  up[v][0] = p;
  upg[v][0] = (v != p) ? merge(gc[v], gc[p]) : gc[v];
  for (int i = 1; i <= LOG; i++) {
    up[v][i] = up[up[v][i - 1]][i - 1];
    upg[v][i] = merge(upg[v][i - 1], upg[up[v][i - 1]][i - 1]);
  }
  for (int to : g[v]) {
    if (to != p) {
      dfs(to, v);
    }
  }
  tout[v] = ++timer;
}
bool upper(int u, int v) { return tin[u] <= tin[v] && tout[u] >= tout[v]; }
node calc(int u, int v) {
  if (u == v) return gc[v];
  node c = gc[v];
  for (int i = LOG; i >= 0; i--) {
    if (!upper(up[v][i], u)) {
      c = merge(c, upg[v][i]);
      v = up[v][i];
    }
  }
  return merge(c, upg[v][0]);
}
int lca(int u, int v) {
  for (int i = LOG; i >= 0; i--) {
    if (!upper(up[u][i], v)) {
      u = up[u][i];
    }
  }
  return up[u][0];
}
int main() {
  cin >> n >> m >> q;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int c;
    cin >> c;
    gc[c].insert(i);
  }
  dfs(1);
  for (int i = 1; i <= q; i++) {
    int u, v, a;
    cin >> u >> v >> a;
    node c;
    if (upper(u, v)) {
      c = calc(u, v);
    } else if (upper(v, u)) {
      c = calc(v, u);
    } else {
      int LCA = lca(u, v);
      c = merge(calc(LCA, u), calc(LCA, v));
    }
    c.print(a);
  }
}